The integrals in Exercises 1–34 converge. Evaluate the integra...

Question

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.

∫₂⁴ dt / [t√(t² − 4)]

Accepted Answer

Welcome back everyone. In this problem we want to evaluate the improper integral of dx divided by x multiplied by the square root of x2 minus 1 between the bones of 1 and 2. Now first, let's rewrite the integral with a limit. So taking our integral. OK We can rewrite it as the limit as B approaches 1. Of the integral from b to 2 of d x divided by x multiplied by the square root of x2 minus 1. So if we can figure out our anti derivative, find the indefinite integral here, then we should be able to use it, figure out the definite integral, and then apply the limit. So let's do that. So solving for the integral, OK, solving for the integral or the indefinite integral of dx divided by x multiplied by the square root of x2 minus 1, it would be best here for us to solve it using trigonometric substitution. So let's set. X to be equal to the sant of theta. In that case, that means the derivative of x with respect to theta would be equal to se theta tan theta. So then dx would be equal to se theta tan theta d theta. And now we have all we need to put it into our. Integral because no, that means the integral then. Is going to become. The integral of sec theta, 1 theta, the theta, OK. Divided by se theta multiplied by the square root of sex square theta minus 1, all in terms of theta, and now we could simplify that to be the integral of sic theta t theta d theta divided by se theta multiplied by the absolute value of tan theta. OK, by identity, that's what the square root of sex squared theta minus 1 would be. Now for the given domain, remember we're eventually evaluating this integral between 1 and 2, OK, and for this domain. OK, then theta would be on the corresponding domain from 0 to 1/3 of pi, and that's pretty important because we would need this to help us evaluate or or to make sense of our tangent term because no, in the first quadrant the tangent function is always non-negative so this would imply then that the absolute value of tan theta would be equal to positive tan theta. I know that would mean. Then, OK, that the integral of sick, we could rewrite this as the integral of sick theta tan theta d theta divided by sick theta multiplied by positive tan theta and if we go ahead and cancel like terms, then really this is going to be the integral. Of d theta, which when we integrate that we get it to be theta plus c and if we think about it since x equals se theta, then theta would be equal to the inverse c can of x plus c. So now that we have this information, that means then. Now our integral between b and 2 of d x divided by x multiplied by the square root of x2 minus 1 is equal to the inverse c can of x, OK, between the bones of b and 2. And if we apply that, OK, then we get it to be equal to the inverse sequence of 2 minus the inverse sant of B, which we could rewrite as the inverse cosine of 12 minus the inverse sant of b. Where we know the inverse cosine of 12 is 1/3 of pi, so this is really 1/3 of pi minus the inverse sequence of B. Know that we have our integral, then we can apply our limit because no this means then that our limit as b approaches 1 of 1/3 of pi minus the inverse 2 of b. Is going to be equal to 1/3 of pi, OK, because here that's a constant. The limit as b approaches 1 of 1/3 of pi would end up being the 1/3 of pi, but for our term the inverse c can of b as b tends to 1. Then that means the inverse sequence of B would tend to the inverse sequant of 1. And now if we apply what we know about the definition of the sequent, then this will be 1/3 of pi minus the inverse cosine of 1 divided by 1 or 1. And the inverse cosine of 1 is 0. So really this is 1/3 of pi minus 0, which equals 1/3 of pi. Therefore, our improper integral of d x divided by x multiplied by the square root of x minus 1 between the bones of 1 and 2 is equal to a third of pi. Thanks a lot for watching everyone. I hope this video helped.

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.∫₂⁴ dt / [t√(t² − 4)]

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The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.
∫₂⁴ dt / [t√(t² − 4)]